Suppose $V$ is finite-dimensional, $U$ is a subspace of $V$, and $S \in L(U,V)$. Prove that there exists an invertible operator $T \in L(V, V)$, such that $T(u) = S(u)$ for every $u \in U$ if and only if $S$ is injective.
I started by taking a basis $B$ of $U$ = {$u_1,...,u_m$}, and expanding it with {$u_{m+1}, ...u_{n}$} such that it is a basis for the entire space $V$.
Then, I was going to use the Identity map $I$, as the map $T$ that I have to prove exists, but this didn't work out. I'm not sure how else to approach this problem.
So you have a basis $\{ u_1, \dots, u_m \}$ for $U$, which can be extended to a basis $\{ u_1, \dots, u_m , u_{m+1}, \dots, u_n \}$ for the whole of $V$.
Now consider the elements $$ v_1 := S(u_1), \ \ \dots \ \ , v_m := S(u_m ).$$ Can you use the injectivity of $S$ to show that $\{ v_1, \dots, v_m \}$ is a linearly independent set, and hence, can be extended to a basis $\{ v_1, \dots, v_m, v_{m + 1}, \dots, v_n \}$ for $V$?
To define $T : V \to V$, the idea is to map $$u_1 \mapsto v_1, \ \ \dots, \ \ u_n \mapsto v_n .$$ Can you use this idea to define a linear map $V \to V$? And can you show that this linear map is bijective (hence invertible), and agrees with $S$ on the subspace $U$?